Asked by Diana
A mass of 12.7 kg is hanging by a steel wire 1.02 m long and 1.12 mm in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is 2.0 1011 N/m2.
Answers
Answered by
drwls
Use Young's modulus (and the length and diameter) to compute the spring constant of the wire, k.
k = (force)/(stretch)
= F/[(F*L)/(pi*r^2*Y)]
= pi*(d^2/4)*Y/L
Then use f = [1/(2*pi)]*sqrt(k/m)
for the oscillation frequency.
k = (force)/(stretch)
= F/[(F*L)/(pi*r^2*Y)]
= pi*(d^2/4)*Y/L
Then use f = [1/(2*pi)]*sqrt(k/m)
for the oscillation frequency.
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